A thin uniform circular disc of mass `M` and radius `R` is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity `omega`. Another disc of the same dimensions but of mass `M//4` is placed gently on the first disc co-axially. show that angular velocity o fthe system is `4 omega//5`.

Explanation

As moment of inertia of a disc throug its centre of mass and perpendicular to its plane is `MR^(2)//2`. Its angular momentum about the given axis when it is rotating with anglar velocity `omega` will be
`L=Iomega=1/2MR^(2)omega`…..i
When another disc of same radius but mass `M//4` is placed on the first the moment of inertia of the system
`I'=1/2MR^(2)+1/2i[M/4]R^(2)=5/8MR^(2)`
And so it `omega'` is the new angular velocity of the system
`L'=I'omega'=5/8MR^(2)omega`.......ii
Now as no external torque has been applied, by conservation of angular momentum we have
`L'=L`, i.e. `1/2 MR^(2)omega'=5/8MR^(2)omega'`
or `omega'=(4/5)omega`